Additional Math – Quadratic Equations – Find p when one root is equal or reciprocal to the other

Formula for Sum and Product of Roots

Sum of Roots:    Product of Roots:
$latex \displaystyle \alpha +\beta =-\frac{b}{a}$       $latex \displaystyle \alpha \times \beta =\frac{c}{a}$

From the Quadratic Equation:  a = 3, b = 2p and c = -p


a) 2 equal roots.

Let the roots be α & α since the roots are equal . Do note that we can also let the roots be β and β.

Sum of Roots:

$latex \displaystyle \alpha +\alpha =-\frac{{2p}}{3}$
$latex \displaystyle 2\alpha =-\frac{{2p}}{3}$
$latex \displaystyle \alpha =-\frac{p}{3}\quad \to (1)$

Product of Roots:

$latex \displaystyle \alpha \times \alpha =\frac{p}{3}$

$latex \displaystyle {{\alpha }^{2}}=\frac{p}{3}\quad \to (2)$

Substitute Equation (1) into (2)

$latex \displaystyle {{\left( {-\frac{p}{3}} \right)}^{2}}=\frac{p}{3}$

$latex \displaystyle \frac{{{{p}^{2}}}}{9}=\frac{p}{3}$

$latex \displaystyle 3{{p}^{2}}=9p$

$latex \displaystyle 3{{p}^{2}}-9p=0$

$latex \displaystyle 3p(p-3)=0$

$latex \displaystyle 3p=0\ or\ p-3=0$

$latex \displaystyle p=0\ or\ p=3$


b) 1 roots is a reciprocal of the other

Let the roots be α & 1/α since the roots are equal . Do note that we can also let the roots be β and 1/β.

Sum of Roots:

$latex \displaystyle \alpha +\frac{1}{\alpha }=-\frac{{2p}}{3}$

$latex \displaystyle \frac{{{{\alpha }^{2}}}}{\alpha }+\frac{1}{\alpha }=-\frac{{2p}}{3}$

$latex \displaystyle \frac{{{{\alpha }^{2}}+1}}{\alpha }=-\frac{{2p}}{3}\quad $

There are two unknown variables α and p, we cannot find the value of p, so we try to find for p using product of roots.

Product of Roots:

$latex \displaystyle \alpha \times \frac{1}{\alpha }=\frac{p}{3}$

$latex \displaystyle 1=\frac{p}{3}\quad $

$latex \displaystyle p=3\ $


Students assume that the roots of the equation are always α and β. I purposely posted this solution to show all of you that the roots does not always have to be so.

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